How can I access the name property of a ProtoField after I declare it?
For example, something along the lines of:
myproto = Proto("myproto", "My Proto")
myproto.fields.foo = ProtoField.int8("myproto.foo", "Foo", base.DEC)
print(myproto.fields.foo.name)
Where I get the output:
Foo
An alternate method that's a bit more terse:
local fieldString = tostring(field)
local i, j = string.find(fieldString, ": .* myproto")
print(string.sub(fieldString, i + 2, j - (1 + string.len("myproto")))
EDIT: Or an even simpler solution that works for any protocol:
local fieldString = tostring(field)
local i, j = string.find(fieldString, ": .* ")
print(string.sub(fieldString, i + 2, j - 1))
Of course the 2nd method only works as long as there are no spaces in the field name. Since that's not necessarily always going to be the case, the 1st method is more robust. Here is the 1st method wrapped up in a function that ought to be usable by any dissector:
-- The field is the field whose name you want to print.
-- The proto is the name of the relevant protocol
function printFieldName(field, protoStr)
local fieldString = tostring(field)
local i, j = string.find(fieldString, ": .* " .. protoStr)
print(string.sub(fieldString, i + 2, j - (1 + string.len(protoStr)))
end
... and here it is in use:
printFieldName(myproto.fields.foo, "myproto")
printFieldName(someproto.fields.bar, "someproto")
Ok, this is janky, and certainly not the 'right' way to do it, but it seems to work.
I discovered this after looking at the output of
print(tostring(myproto.fields.foo))
This seems to spit out the value of each of the members of ProtoField, but I couldn't figure out the correct way to access them. So, instead, I decided to parse the string. This function will return 'Foo', but could be adapted to return the other fields as well.
function getname(field)
--First, convert the field into a string
--this is going to result in a long string with
--a bunch of info we dont need
local fieldString= tostring(field)
-- fieldString looks like:
-- ProtoField(188403): Foo myproto.foo base.DEC 0000000000000000 00000000 (null)
--Split the string on '.' characters
a,b=fieldString:match"([^.]*).(.*)"
--Split the first half of the previous result (a) on ':' characters
a,b=a:match"([^.]*):(.*)"
--At this point, b will equal " Foo myproto"
--and we want to strip out that abreviation "abvr" part
--Count the number of times spaces occur in the string
local spaceCount = select(2, string.gsub(b, " ", ""))
--Declare a counter
local counter = 0
--Declare the name we are going to return
local constructedName = ''
--Step though each word in (b) separated by spaces
for word in b:gmatch("%w+") do
--If we hav reached the last space, go ahead and return
if counter == spaceCount-1 then
return constructedName
end
--Add the current word to our name
constructedName = constructedName .. word .. " "
--Increment counter
counter = counter+1
end
end
Related
for my code I want all numbers from a dictionary under 70 to be deleted, I'm unsure of how to specify this and I need it to also delete the associated name with that number as well, either that or only diplay numbers that are 70 or above.
Below is the code that I have in it's entirety:
name = []
number =[]
name_grade = {}
counter = 0
counter_bool= True
num_loop = True
while counter_bool:
stu = int(input("please enter the number of students: "))
if stu < 2:
print("value is too low, try again")
continue
else:
break
while counter != stu:
name_inp = str(input("Enter your name: "))
while num_loop:
number_inp = int(input("Enter your number: "))
if number_inp < 0 or number_inp > 100:
print("The value is too high or too low, please enter a number between 0 and 100.")
continue
else:
break
name_grade[name_inp] = number_inp
name.append(name_inp)
number.append(number_inp)
counter += 1
print(name_grade)
sorted_numbers = sorted(name_grade.items(), key= lambda x:x[1])
print(sorted_numbers)
if number > 70:
resorted_numbers = number < 70
print(resorted numbers)
how would I go about this?
Also if it's also not too much trouble could someone explain in detail about dictionary keys and how the lambda function I've used works? I got help but I would prefer to know the small details on how it's applied and formatted but don't worry if it's a pain to explain.
You can just iterate over the dictionary and filter for values less than 70:
resorted_numbers = {k:v for k,v in name_grade.items() if v<70}
dict.items method returns a list of key-value tuple pairs of a dictionary, so the lambda function is telling the sorted function to sort by the second element in each tuple.
I am wondering if its possible to search in the database with the given scrambled words.
I have a mobs table in database and it holds the name of the monster names
If given monster name is A Golden Dregon or A Golden Dfigon or A Gelden Dragon I want it to find A Golden Dragon or with the matches that close to it from database. Usually one or two letters at max is given like this as scrambled.
Is that possible with just SQL queries? Or should I build the query by parsing the given monster name?
I am using LUA for the code side.
I have come to know this search type as a fuzzy search. I mainly program in JS and use fuse.js all the time for this kind of problem.
Fuzzy Searches are based on the Levenshtein algorithm that rate the distance of two strings. When you have this distance value you can sort or drop elements from a list based on the score.
I found the algorithm in lua here.
function levenshtein(s, t)
local s, t = tostring(s), tostring(t)
if type(s) == 'string' and type(t) == 'string' then
local m, n, d = #s, #t, {}
for i = 0, m do d[i] = { [0] = i } end
for j = 1, n do d[0][j] = j end
for i = 1, m do
for j = 1, n do
local cost = s:sub(i,i) == t:sub(j,j) and 0 or 1
d[i][j] = math.min(d[i-1][j]+1, d[i][j-1]+1, d[i-1][j-1]+cost)
end
end
return d[m][n]
end
end
As explained in the site you compare two strings like so and get a score based on the distance of them, then sort or drop the items being search based on the scores given. As this is CPU expensive I would suggest caching or use a memoize function to store common mistakes.
levenshtein('referrer', 'referrer') -- zero distance
>>> 0
levenshtein('referrer', 'referer') -- distance of one character
>>> 1
levenshtein('random', 'strings') -- random big distance
>>> 6
Got a simple version of it working in lua here I must say lua is an easy language to pick up and start coding with.
local monsters = {'A Golden Dragon', 'Goblins', 'Bunny', 'Dragoon'}
function levenshtein(s, t)
local s, t = tostring(s), tostring(t)
if type(s) == 'string' and type(t) == 'string' then
local m, n, d = #s, #t, {}
for i = 0, m do d[i] = { [0] = i } end
for j = 1, n do d[0][j] = j end
for i = 1, m do
for j = 1, n do
local cost = s:sub(i,i) == t:sub(j,j) and 0 or 1
d[i][j] = math.min(d[i-1][j]+1, d[i][j-1]+1, d[i-1][j-1]+cost)
end
end
return d[m][n]
end
end
--Fuzzy Search Returns the Best Match in a list
function fuzzySearch(list, searchText)
local bestMatch = nil;
local lowestScore = nil;
for i = 1, #list do
local score = levenshtein(list[i], searchText)
if lowestScore == nil or score < lowestScore then
bestMatch = list[i]
lowestScore = score
end
end
return bestMatch
end
print ( fuzzySearch(monsters, 'golen dragggon') )
print ( fuzzySearch(monsters, 'A Golden Dfigon') )
print ( fuzzySearch(monsters, 'A Gelden Dragon') )
print ( fuzzySearch(monsters, 'Dragooon') ) --should be Dragoon
print ( fuzzySearch(monsters, 'Funny') ) --should be Bunny
print ( fuzzySearch(monsters, 'Gob') ) --should be Goblins
Output
A Golden Dragon
A Golden Dragon
A Golden Dragon
Dragoon
Bunny
Goblins
For SQL
You can try to do this same algorithm in T-SQL as talked about here.
In SQLlite there is an extension called editdist3 which also uses this algorithm the docs are here.
I would be hard to compensate for all the different one and two letter scrambled combinations, but you could create a lua table of common misspellings of "A Golden Dragon" check if it is in the table. I have never used lua before but here is my best try at some sample code:
local mob_name = "A Golden Dregon"--you could do something like, input("Enter mob name:")
local scrambled_dragon_names = {"A Golden Dregon", "A Golden Dfigon", "A Gelden Dragon"}
for _,v in pairs(scrambled_dragon_names) do
if v == mob_name then
mob_name = "A Golden Dragon"
break
end
end
I really hope I have helped!
P.S. If you have anymore questions go ahead and comment and I will try to answer ASAP.
You will have to parse the given monster name to some extent, by making assumptions about how badly it is misspelled. For example, if the user supplied the name
b fulden gorgon
There is no way in hell you can get to "A Golden Dragon". However, if you assume that the user will always get the first and last letters of every word correctly, then you could parse the words in the given name to get the first and last letters of each word, which would give you
"A", "G" "n", "D" "n"
Then you could use the LIKE operator in your query, like so:
SELECT * FROM mobs WHERE monster_name LIKE 'A G%n D%n';
The main point here is what assumptions you make about the misspelling. The closer you can narrow it down, the better your query results will be.
Using HtmlProvider to access a web-based table sometimes returns a fraction as a string (correct) and, at other times, returns a DateTime (incorrect).
What am I missing?
module Test =
open FSharp.Data
let [<Literal>] url = "https://www.example.com/fractions"
type profile = HtmlProvider<url>
let profile = profile.Load(url)
let [<Literal>] resultFile = #"C:\temp\data\Profile.csv"
let CsvResult =
do
use writer = new StreamWriter(resultFile, false)
writer.WriteLine "\"Date\";\"Fraction\""
for row in profile.Tables.Table1.Rows do
"\"" + row.``Date``.ToString() + "\"" + ";" |> writer.Write
"\"" + row.``Fraction``.ToString() + "\"" + ";" |> writer.WriteLine
writer.Close
let csvResult = CsvResult
Without seeing sample data I can't be 100% certain, but I'm guessing that it's parsing fractions as dates if the numbers involved would be valid dates in the culture you're using: e.g., 1/4 would be a valid date in any culture that uses / as a separator, and would be treated either as April 1st or as January 4th, depending on which parsing culture your system defaults to.
Other type providers in FSharp.Data (such as the CSV type provideryou could ) allow you to configure how each column will be parsed, but that's not an option the HTML type provider gives you. (Which is a bit of a missing feature, of course). But since the HTML type provider does allow you to specify the culture info for datetime and number parsing, one way you might be able to work around this is specify a culture that does not use / as a separator (but still uses . as a decimal point, since otherwise if the HTML you're parsing has numbers written like 1,000 for one thousand, that could be interpreted as 1). One such culture is the en-IN culture ("English (India)"), where the date separator is - and the decimal point is ..
So try passing Culture=System.Globalization.CultureInfo.GetCultureInfo("en-IN") in your HtmlProvider options, and see if that helps it stop treating fractions as dates.
The following combination of functions worked:
// http://www.fssnip.net/29/title/Regular-expression-active-pattern
module Solution =
open System
open System.Text.RegularExpressions
open FSharp.Data
let (|Regex|_|) pattern input =
let m = Regex.Match(input, pattern)
if m.Success then Some(List.tail [ for g in m.Groups -> g.Value ])
else None
let ptrnFraction = #"^([0-9]?[0-9]?)(\/)([0-9]?[0-9]?)$"
let ptrnDateTime = #"(\d{2})\/(\d{2})\/(\d{4}) (\d{2}):(\d{2}):(\d{2})"
let ToFraction input =
match input with
| Regex ptrnFraction [ numerator; operator; denominator ] ->
(numerator + operator + denominator).ToString()
| Regex ptrnDateTime [ day; month; year; hours; minutes; seconds ] ->
(day + "/" + month).ToString()
| _ -> "Not valid!"
let dtInput = #"05/09/2017 00:00:00"
let frcInput = #"13/20"
let outDate = ToFraction dtInput
printfn "Out Date: %s" outDate
let outFraction = ToFraction frcInput
printfn "Out Fraction: %s" outFraction
//Output:> Out Date: 05/09 Out Fraction: 13/20
Thus, I was able to replace:
"\"" + row.``Fraction``.ToString() + "\"" + ";" |> writer.WriteLine
with:
"\"" + ToFraction(row.``Fraction``.ToString()) + "\"" + ";" |> writer.Write
Thanks to #rmunn for the clarity of his explanations and the benefit of his expertise.
Say I have a string.
"poop"
I want to change "poop" to "peep".
In fact, I also want all of the o's in poop to change to e's for any word I put in.
Here's my attempt to do the above.
def getword():
x = (input("Please enter a word."))
return x
def main():
y = getword()
for i in range (len(y)):
if y[i] == "o":
y = y[:i] + "e"
print (y)
main()
As you can see, when you run it, it doesn't amount to what I want. Here is my expected output.
Enter a word.
>>> brother
brether
Something like this. I need to do it using slicing. I just don't know how.
Please keep your answer simple, since I'm somewhat new to Python. Thanks!
This uses slicing (but keep in mind that slicing is not the best way to do it):
def f(s):
for x in range(len(s)):
if s[x] == 'o':
s = s[:x]+'e'+s[x+1:]
return s
Strings in python are non-mutable, which means that you can't just swap out letters in a string, you would need to create a whole new string and concatenate letters on one-by-one
def getword():
x = (input("Please enter a word."))
return x
def main():
y = getword()
output = ''
for i in range(len(y)):
if y[i] == "o":
output = output + 'e'
else:
output = output + y[i]
print(output)
main()
I'll help you this once, but you should know that stack overflow is not a homework help site. You should be figuring these things out on your own to get the full educational experience.
EDIT
Using slicing, I suppose you could do:
def getword():
x = (input("Please enter a word."))
return x
def main():
y = getword()
output = '' # String variable to hold the output string. Starts empty
slice_start = 0 # Keeps track of what we have already added to the output. Starts at 0
for i in range(len(y) - 1): # Scan through all but the last character
if y[i] == "o": # If character is 'o'
output = output + y[slice_start:i] + 'e' # then add all the previous characters to the output string, and an e character to replace the o
slice_start = i + 1 # Increment the index to start the slice at to be the letter immediately after the 'o'
output = output + y[slice_start:-1] # Add the rest of the characters to output string from the last occurrence of an 'o' to the end of the string
if y[-1] == 'o': # We still haven't checked the last character, so check if its an 'o'
output = output + 'e' # If it is, add an 'e' instead to output
else:
output = output + y[-1] # Otherwise just add the character as-is
print(output)
main()
Comments should explain what is going on. I'm not sure if this is the most efficient or best way to do it (which really shouldn't matter, since slicing is a terribly inefficient way to do this anyways), just the first thing I hacked together that uses slicing.
EDIT Yeah... Ourous's solution is much more elegant
Can slicing even be used in this situation??
The only probable solution I think would work, as MirekE stated, is y.replace("o","e").
Given a vector, A=[1,2,10,10,10,1], [m,k]=max(A) returns k=3.
How to get the last index of the maximum element? In this example, I want to get 5.
The most succinct way I can think of is: [m,k]=max(A($:-1:1))
Is there a better way or does scilab provide a parameter to do this? Reversing a large vertex is not a good idea in any way.
Use the find command
You can use the find command to get all indices of the maximum:
indices = find(A==max(A))
last = max(indices)
Implement it yourself
Or if you want a single pass, you can implement it yourself:
//Create a c-function with your wanted behaviour
f1=['void max_array(int in_array[],int* in_num_elements,int *out_max, int *out_index)'
'{'
'int i;'
'*out_max=in_array[0];'
'*out_index=-1;'
'for (i=0; i<*in_num_elements; i++)'
'{'
'if(in_array[i]>=*out_max)'
'{'
'*out_max=in_array[i];'
'*out_index=i;'
'}'
'}'
'}'];
//Place it in a file in the current directory
mputl(f1,'max_array.c')
//Create the shared library (a gateway, a Makefile and a loader are
//generated.
ilib_for_link('max_array','max_array.c',[],"c")
//Load the library
exec loader.sce
//Create wrapper for readability
function [m,k] = last_max(vector)
[m, k] = call('max_array', vector, 1,'i', length(vector),2,'i', 'out',[1,1],3,'i',[1,1],4,'i');
// Because c is zero-indexed add 1
k = k + 1
endfunction
//Your data
A=[1,2,10,10,10,1];
//Call function on your data
[m,k] = last_max(A)
disp("Max value is " + string(m) + " at index " + string(k) )